宸茬煡鍏充簬x镄勬柟绋媥虏+2(a-1)x+a虏-7a-4=0

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宸茬煡鍏充簬x镄勬柟绋媥虏+2(a-1)x+a虏-7a-4=0
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宸茬煡鍏充簬x镄勬柟绋媥虏+2(a-1)x+a虏-7a-4=0
宸茬煡鍏充簬x镄勬柟绋媥虏+2(a-1)x+a虏-7a-4=0

宸茬煡鍏充簬x镄勬柟绋媥虏+2(a-1)x+a虏-7a-4=0
由题意可知:Δ=4(a-1)²-4(a²-7a-4)≥0
即:4a²-8a+4-(4a²-28a-16)≥0
则有:20a+20≥0
解得:a≥-1
又根据韦达定理可得:
x1+x2=-2(a-1),x1*x2=a²-7a-4
因为x1*x2-3x1-3x2-2=0,那么:
x1*x2-3(x1+x2)-2=0
所以:a²-7a-4+6(a-1)-2=0
a²-a-12=0
(a+3)(a-4)=0
由于a≥-1,所以解得:a=4(另a=-3不合题意,舍去)
所以:
[1+ 4/(a²-4)]*(a+2)/a
=a²/[(a+2)(a-2)] *(a+2)/a
=a/(a-2)
=4/(4-2)
=2